Composite Bézier curve

Smooth joining

A commonly desired property of splines is for them to join their individual curves together with a specified level of parametric or geometric continuity. While individual curves in the spline are fully ${\displaystyle C^{\mathrm{\infty }}}$ continuous within their own interval, there is always some amount of discontinuity where different curves meet.

The Bézier spline is fairly unique in that it's one of the few splines that doesn't guarantee any higher degree of continuity than ${\displaystyle C^0}$. It is, however, possible to arrange control points to guarantee various levels of continuity across joins, though this can come at a loss of local control if the constraint is too strict for the given degree of the Bézier spline.

Smoothly joining cubic Béziers

Given two cubic Bézier curves with control points ${\displaystyle [{\mathbf{P}}_0,{\mathbf{P}}_1,{\mathbf{P}}_2,{\mathbf{P}}_3]}$ and ${\displaystyle [{\mathbf{P}}_3,{\mathbf{P}}_4,{\mathbf{P}}_5,{\mathbf{P}}_6]}$ respectively, the constraints for ensuring continuity at ${\displaystyle {\mathbf{P}}_3}$ can be defined as follows:

• ${\displaystyle C^0/G^0}$ (positional continuity) requires that they meet at the same point, which all polybéziers do by definition. In this example, the shared point is ${\displaystyle {\mathbf{P}}_3}$
• ${\displaystyle C^1}$ (velocity continuity) requires the neighboring control points around the join to be mirrors of each other. In other words, they must follow the constraint of ${\displaystyle {\mathbf{P}}_4=2{\mathbf{P}}_3-{\mathbf{P}}_2}$
• ${\displaystyle G^1}$ (tangent continuity) requires the neighboring control points to be collinear with the join. This is less strict than ${\displaystyle C^1}$ continuity, leaving an extra degree of freedom which can be parameterized using a scalar ${\displaystyle {\beta }_1}$. The constraint can then be expressed by ${\displaystyle {\mathbf{P}}_4={\mathbf{P}}_3+({\mathbf{P}}_3-{\mathbf{P}}_2){\beta }_1}$

While the following continuity constraints are possible, they are rarely used with cubic polybéziers, as other splines like the B-spline or the β-spline will naturally handle higher constraints without loss of local control.

• ${\displaystyle C^2}$ (acceleration continuity) is constrained by ${\displaystyle {\mathbf{P}}_5={\mathbf{P}}_1+4({\mathbf{P}}_3-{\mathbf{P}}_2)}$. However, applying this constraint across an entire cubic Bézier spline will cause a cascading loss of local control over the tangent points. The curve will still pass through every third point in the spline, but control over its shape will be lost. In order to achieve ${\displaystyle C^2}$ continuity using cubic curves, it's recommended to use a cubic uniform B-spline instead, as it ensures ${\displaystyle C^2}$ continuity without loss of local control, at the expense of no longer being guaranteed to pass through specific points
• ${\displaystyle G^2}$ (curvature continuity) is constrained by ${\displaystyle {\mathbf{P}}_5={\mathbf{P}}_3+({\mathbf{P}}_3-{\mathbf{P}}_2)(2{\beta }_1+{\beta }_1^2+{\beta }_2/2)+({\mathbf{P}}_1-{\mathbf{P}}_2){\beta }_1^2}$, leaving two degrees of freedom compared to ${\displaystyle C^2}$, in the form of two scalars ${\displaystyle {\beta }_1}$ and ${\displaystyle {\beta }_2}$. Higher degrees of geometric continuity is possible, though they get increasingly complex
• ${\displaystyle C^3}$ (jolt continuity) is constrained by ${\displaystyle {\mathbf{P}}_6={\mathbf{P}}_3+({\mathbf{P}}_3-{\mathbf{P}}_0)+6({\mathbf{P}}_1-{\mathbf{P}}_2+{\mathbf{P}}_3-{\mathbf{P}}_2)}$. Applying this constraint to the cubic Bézier spline will cause a complete loss of local control, as the entire spline is now fully constrained and defined by the first curve's control points. In fact, it is arguably no longer a spline, as its shape is now equivalent to extrapolating the first curve indefinitely, making it not only ${\displaystyle C^3}$ continuous, but ${\displaystyle C^{\mathrm{\infty }}}$, as joins between separate curves no longer exist

Approximating circular arcs

In case circular arc primitives are not supported in a particular environment, they may be approximated by Bézier curves. Commonly, eight quadratic segments or four cubic segments are used to approximate a circle. It is desirable to find the length ${\displaystyle \mathbf{k}}$ of control points which result in the least approximation error for a given number of cubic segments.

Using four curves

Considering only the 90-degree unit-circular arc in the first quadrant, we define the endpoints ${\displaystyle \mathbf{A}}$ and ${\displaystyle \mathbf{B}}$ with control points ${\displaystyle {\mathbf{A}}^{\prime }}$ and ${\displaystyle {\mathbf{B}}^{\prime }}$, respectively, as:

${\displaystyle \begin{array}{cc}\mathbf{A}& =[0,1]\\ {\mathbf{A}}^{\prime }& =[\mathbf{k},1]\\ {\mathbf{B}}^{\prime }& =[1,\mathbf{k}]\\ \mathbf{B}& =[1,0]\end{array}}$

From the definition of the cubic Bézier curve, we have:

${\displaystyle \mathbf{C}(t)=(1-t{)}^3\mathbf{A}+3(1-t{)}^{2}t{\mathbf{A}}^{\prime }+3(1-t)t^2{\mathbf{B}}^{\prime }+t^3\mathbf{B}}$

With the point ${\displaystyle \mathbf{C}(t=0.5)}$ as the midpoint of the arc, we may write the following two equations:

${\displaystyle \begin{array}{ccc}\mathbf{C}& =\frac{1}{8}\mathbf{A}+\frac{3}{8}{\mathbf{A}}^{\prime }+\frac{3}{8}{\mathbf{B}}^{\prime }+\frac{1}{8}\mathbf{B}& \text{(by substitution into cubic Bézier definition)}\\ \mathbf{C}& =[\sqrt{2}/2,\sqrt{2}/2]& \text{(by the definition of a circle)}\end{array}}$

Solving these equations for the x-coordinate (and identically for the y-coordinate) yields:

${\displaystyle \frac{0}{8}\mathbf{+}\frac{3}{8}\mathbf{k}+\frac{3}{8}+\frac{1}{8}=\sqrt{2}/2}$

${\displaystyle \mathbf{k}=\frac{4}{3}(\sqrt{2}-1)\approx 0.5522847498}$

Error analysis

Note however that the resulting Bézier curve is entirely outside the circle, with a maximum deviation of the radius of about 0.00027. For the rest of the discussion, the radial drift for a given t is ${\displaystyle \sqrt{x(t{)}^2+y(t{)}^2}-1}$.

Unattributed correction: Adding a small correction to intermediate points such as

${\displaystyle \begin{array}{cc}{\mathbf{A}}^{\prime }& =[\mathbf{k}+0.0009,1-0.00103]\\ {\mathbf{B}}^{\prime }& =[1-0.00103,\mathbf{k}+0.0009],\end{array}}$

the magnitude of the radius deviation to 1 is reduced by a factor of about 3, to 0.000068 (at the expense of the derivability of the approximated circle curve at endpoints).

Ellsworth's three-parameter correction: Instead of solving for a single k, solve for ${\displaystyle \mathbf{A}=[0,a],}$ ${\displaystyle {\mathbf{A}}^{\prime }=[b,c],}$ ${\displaystyle {\mathbf{B}}^{\prime }=[c,b],}$ ${\displaystyle \mathbf{B}=[a,0].}$ Minimize the maximum radial drift, which is $f(a,b,c)=max(\left|d(t)\right|,0\le t\le 1)$. Solving numerically gives a = 1.00005507808, b = 0.55342925736, and c = 0.99873327689. The maximum radial drift is 0.000055.

Ellsworth's least-square correction: Solve for the k that minimizes ${\int }_0^1{\left(d(t)\right)}^{2}dt$. Numerically ${\displaystyle k\approx 0.55197038140}$. The maximal radial drift is 0.0002, for a 1.9% reduction in the total squared radial drift compared to the three-parameter case.

General case

We may approximate a unit circle of radius from an arbitrary number of cubic Bézier curves. Let the arc start at point ${\displaystyle \mathbf{A}}$ and end at point ${\displaystyle \mathbf{B}}$, placed at equal distances above and below the x-axis, spanning an arc of angle ${\displaystyle \theta =2\varphi \le <!--\; -->=\frac{\pi }{2}}$:

${\displaystyle \begin{array}{cc}\mathbf{A}& =[\cos (\varphi ),\sin (\varphi )]\\ \mathbf{B}& =[\cos (\varphi ),-\sin (\varphi )]=[{\mathbf{A}}_x,-{\mathbf{A}}_y]\end{array}}$

After solving for the curve starting and ending at the given points, the arc's midpoint occurring at ${\displaystyle t=0.5}$, and the curve's end point tangents being the same as the circles, the control points may be written as:

${\displaystyle \begin{array}{ccc}{\mathbf{A}}^{\prime }& =[\frac{4-{\mathbf{A}}_x}{3},\frac{(1-{\mathbf{A}}_x)(3-{\mathbf{A}}_x)}{3{\mathbf{A}}_y}]& =[\frac{4-\cos (\varphi )}{3},\frac{(1-\cos (\varphi ))(3-\cos (\varphi ))}{3\sin (\varphi )}]\\ {\mathbf{B}}^{\prime }& =[{{\mathbf{A}}^{\prime }}_x,-{{\mathbf{A}}^{\prime }}_y]& =[\frac{4-\cos (\varphi )}{3},\frac{(1-\cos (\varphi ))(3-\cos (\varphi ))}{-3\sin (\varphi )}]\end{array}}$

Multiplying the values provide general solutions for arcs of any radius. Using rotation matrices allows the drawing of any angle.

The constraints of Ellsworth's solutions can be applied to the general case and numerically solved as well.

Examples

Other approaches

For Bézier curves of degree n > 1, Ahn (2019) provide an ${\displaystyle G^2}$ approximation for any number of segments. The Hausdorff distance as well as the aforementioned "radial drift" show 2n-1 extreme points and Ahn's approximation is, as of the time of publication, the one with the smallest maximum Hausdorff distance. Explicit, analytical solutions are provided for the quadratic, cubic, and quartic cases.

Applying affine transformation to the constraints allows the drawing of ellipses and elliptical arcs.

Fonts

TrueType fonts use composite Béziers composed of quadratic Bézier curves (2nd order curves) while most PostScript Type 1 and OpenType CFF use cubic (3rd order) curves. To describe a typical type design as a computer font to any given accuracy, 3rd order Béziers require less data than 2nd order Béziers; and these in turn require less data than a series of straight lines. This is true even though any one straight line segment requires less data than any one segment of a parabola; and that parabolic segment in turn requires less data than any one segment of a 3rd order curve.